Schrödinger s Wave Equation

Transcription

1 Chapter 4 Schrödinger s Wave Equation 4.1 Introduction We have introduced the idea of a wavefunction ψ(x) that incorporates the wave nature of a particle, and we have introduced the idea that the square of the magnitude gives the probability density for locating a particle, i.e., ρ(x) = ψ(x) 2. We have also presented Schrödinger s equation h2 d 2 ψ(x) 2m dx 2 + U(x) ψ(x) = E ψ(x). (4.1) The solutions of this differential equation correspond to special-case wavefunctions that represent states of definite energy. These special-case wave functions have analogs in classical standing waves. For the bound (confined) physical systems that we will study, the special-case wavefunctions for a given potential will be part of a discrete set, which I label ψ n (x). 4.2 More Practice with Differential Equations In PHYS 212E we have already identified solutions to simple differential equations. This section simply provides more practice. Let s start with a simple differential equation whose solution is function f(x): df(x) = 4. (4.2) dx The solution of this equation is not a number or a variable. Rather, it is an expression for the function f(x) that, when plugged into Eq. (4.2), makes the left and right sides of the equation equal. The question you should ask yourself is What function is there, that if I take one derivative with respect to x, gives the answer 4. In this simple example, it should be clear that the solution is y(x) = 4x + constant. (4.3) 23

2 24 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION The constant in this example can be determined by knowing the value of the function at one point, e.g., if y(1) = 5, then the constant is 9. In working with differential equations in this course we will exploit one of three approaches: The solution will be obvious to anyone who has studied basic calculus (like the example above); You will be guided to guess a solution that contains some undetermined constants. Plugging the guess into the differential equation will provide conditions on some of the constants that make the proposed guess a solution. This process might show that it is impossible to adjust the constants so that the guess can be a solution in this case, the guess is just wrong. You will check a complete given function to verify that it is, indeed, a solution of the differential equation. Example 1. Testing a solution of a differential equation Consider the differential equation d 2 y = 9y. (4.4) dx2 Test whether the function y(x) = Ae cx is a solution of this equation. If it can be a solution, what conditions are placed on the constants A and c? Solution: The first two derivatives of y are dy dx = caex and d 2 y dx 2 = c2 Ae cx. (4.5) Using this expression for the second derivative on the left-hand side of Eq. (4.4), and the original expression for y(x) on the right-hand side, we test the solution (the fact that it is a test is indicated with the question mark over the equals sign): Canceling common factors leaves c 2 Ae cx? = 9 Ae cx. (4.6) c 2? = 9. (4.7)

3 4.2. MORE PRACTICE WITH DIFFERENTIAL EQUATIONS 25 This last statement will be true if c = ±3, giving two possible solutions to this differential equation. Our test has yielded no conditions on the constant A, so the possible solutions are y 1 (x) = A 1 e 3x and y 2 (x) = A 2 e 3x (4.8) You can check that adding these two solutions yields another function that is also a solution: y(x) = A 1 e 3x + A 2 e 3x. (4.9) We need additional information to determine values for A 1 and A 2. Example 2. equation Failed test of a proposed solution of a differential Test the trial solution y(x) = Bx 2 to see if it can be a solution of the differential equation d2 y(x) dx 2 + dy dy = A y(x). Solution: Calculate the derivatives of our trial solution: dy dx = 2Bx and d 2 y = 2B. (4.10) dx2 Next, substitute the derivatives and the function into the differential equation, giving and then simplify, giving 2B + 2Bx? = A Bx 2, (4.11) Ax 2 2x + 2 = 0. (4.12) In this equation there is no choice of A that will make this true for all values of x, so y(x) can t be a solution. 1 (For example, if A = 1 2, the equation will be true for x = 2, as you can find using the quadratic equation, but it won t be true for any other values of x.) In general, a polynomial like Ax 3 + Bx 2 + Cx + D = 0 (4.13) can only be true for all values of x if A, B, C, and D are all zero. 1 Choosing B = 0 would work, but then you d have y(x) = 0, which isn t a very interesting solution, and could never be a wavefunction.

4 26 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION In lab you will construct solutions to Schrödinger s equation without the guess-and-check method using a numerical iteration method and a computer, that is similar in spirit to the numerical integration of Newton s second law that you did in PHYS 211 and in a Python lab earlier this semester. There are many more advanced techniques for finding solutions to differential equations like Schrödinger s equation from scratch that you might learn in a math class, but we won t use them in this course. 4.3 Some Examples of Solutions of Schrödinger s Equation In this section we will investigate at some specific special-case solutions of Schrödinger s equation Particle in a box [You should have already worked out the wavefunctions ψ n (x) and energies E n for the special-case wavefunctions for the a particle in the infinite square potential when you completed Problem 3.(15), but I include this discussion as a reference.] To demonstrate how we use Schrödinger s equation to derive wavefunctions, let s consider the simplest example of a particle in a box (or infinite square well potential) introduced in Section 3.3 of the previous chapter; for a quick reminder, see Fig. 3.1 on page 5. Within the region 0 < x < L the potential U(x) is zero, so Schrödinger s equation, Eq. (4.1), simplifies to h2 d 2 ψ(x) 2m dx 2 = E ψ(x), (4.14) and a slight rearrangement gives d 2 ( ) ψ(x) 2mE dx 2 = h 2 ψ(x). (4.15) Note that the constant within the parentheses on the right is a positive number. For convenience we call this constant k 2, where k 2 = 2mE h 2 > 0. (4.16) In terms of this constant Schrödinger s equation becomes d 2 ψ(x) dx 2 = k 2 ψ(x). (4.17)

5 4.3. SOLUTIONS OF SCHRÖDINGER S EQUATION 27 Our job is to determine the function ψ(x) that is a solution of this differential equation. The question is: What function has a second derivative that is equal to the function itself times a negative number? There are only two real functions that have this property: A sin(kx) and B cos(kx). If it isn t obvious that these are solutions, use the guess-and-check method introduced earlier. (You might think that Ce kx would be a solution, but it isn t. Again, if it isn t obvious that this isn t a solution, use the guessand-check method, and see why this fails as a solution.) So inside the box, the solution must be of the form or or most generally ψ(x) = A sin(kx), (4.18) ψ(x) = B cos(kx), (4.19) ψ(x) = A sin(kx) + B cos(kx), (4.20) where the constants A, B, and k have yet to be determined. Outside of the interval 0 < x < L the potential U(x) goes to infinity, so Schrödinger s equation becomes h2 d 2 ψ(x) 2m dx 2 + ψ(x) = Eψ(x). (4.21) The only way to satisfy this equation is if ψ(x) = 0 in this region, which is consistent with the idea that there is no probability that the particle can penetrate into the infinite potential barriers at x = 0 and x = L. Specifically, this tells us that ψ(0) = 0 and ψ(l) = 0. (4.22) Because probability densities must be continuous functions, the wavefunction on the inside of the box must match up at the edges with the wavefunction on the outside of the box. Therefore we can use these conditions in Eq. (4.20) to learn more about the undetermined constants k, A, and B that appear in the expression for ψ inside the box. The first condition, ψ(0) = 0, implies that ψ(0) = A sin(0) + B cos(0) = 0. (4.23) This fact imposes the requirement that B = 0 (because sin(0) = 0 and cos(0) = 1). The second condition, ψ(l) = 0, imposes the requirement A sin(kl) = 0. (4.24) There are many values of k that will make this true: all values that make kl an integer multiple of π. This means that k n L = nπ. (4.25)

6 28 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION Rearranging gives k n = nπ. (n = 1, 2, 3,... ) (4.26) L (We have thrown out the k = 0 solution because that would correspond to a wavefunction ψ(x) = 0 everywhere, which corresponds to a particle that has zero probability of being found anywhere, i.e., the particle wouldn t exist.) Recalling the definition of k in Eq. (4.16), we see that a discrete set of k s implies a discrete set of possible energies for the particle in a box given by E n = n 2 π2 h 2 2mL 2 = n2 h 2 8mL 2 (n = 1, 2, 3,... ), (4.27) The wavefunctions associated with each of these energies are ( nπ ) ψ n (x) = A sin L x. (4.28) The last thing to be determined is the value of the constant A. This comes from the normalization condition 1 = = L 0 ψ(x) 2 dx A 2 sin 2 ( nπx L ) dx = A 2 L 2, (4.29) so 2 A = L. (4.30) This completes the formal process for determining the special-case wavefunctions for the particle-in-a-box problem. It s really the same process of fitting de Broglie waves into the box that we did before, but Schrödinger s equation and the tools of differential equations are generalizable to more complicated physical systems. Example 3. Electron trapped in a 1-D nanotube Carbon atoms can be bonded into a cylindrical arrangement called a carbon nanotube. Carbon nanotubes can be fabricated with lengthto-diameter ratios that are extremely large. These nanotubes have a broad range of applications including uses as conducting nano-wires and mechanical scaffolding for growing new bone cells. A carbon nanotube of length L can be approximated as an infinite square well of width L. Consider a particular carbon nanotube of

7 4.3. SOLUTIONS OF SCHRÖDINGER S EQUATION 29 length L = 10.0 nm and negligible diameter. Calculate the ground state energy for an electron trapped inside of this carbon nanotube. Solution: Approximating the carbon nanotube as an infinite square well, the possible allowed energies for the electron are given by Eq. (4.27). For the ground state energy, where n = 1, the energy is given as E 1 = (1) 2 h 2 8m e L 2. (4.31) In order to keep everything in terms of nanometers, I multiply the numerator and denominator by c 2, giving E 1 = (hc)2 8(m e c 2 )L 2. (4.32) Inserting values for the mass of the electron, m e c 2 = 0.51 MeV, length of the nanotube L = 10 nm, and hc = 1240 ev nm we get E 1 = (1240 ev nm) ev (10 nm) 2 = 3.8 mev = J (4.33) Particle in a more complicated box Consider a particle confined to move in one dimension by the potential illustrated on the left side of Fig. 4.1: 0 for 0 < x < L 2 U(x) = 50 h2 for L 8mL 2 2 x < L. (4.34) elsewhere The classical motion of a particle in this more complicated box depends on its total energy. If the energy is below 50h 2 /(8mL 2 ) the particle is confined to travel back and forth between x = 0 and x = L/2; when the energy is greater than this threshold, it travels between the two end walls at x = 0 and x = L, but the particle has less kinetic energy in the right half of the box than it does in the left, and therefore the particle moves slower on the right than it does on the left. Aspects of the classical behavior are evident in the illustrated special-case wavefunctions in Fig. 4.1, but there are some radical new features. At high enough energies, i.e., E > 50h 2 /(8mL 2 ), (see graphs of φ 5 (x) and φ 6 (x)), the wavefunction is sinusoidal in both halves of the box, but the wavelength is shorter in the left half of the box. According to the de Broglie hypothesis this corresponds to a particle with a higher momentum in the left half. At

8 30 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION units: h 2 8mL U E 3 E 2 E 1 0 L/2 L E 6 E 5 x φ2(x) (units: 1/ L) φ5(x) (units: 1/ L) φ6(x) (units: 1/ L) x (units: L) x (units: L) x (units: L) φ1(x) (units: 1/ L) x (units: L) Figure 4.1: Particle-in-a-more-complicated-box.

9 4.3. SOLUTIONS OF SCHRÖDINGER S EQUATION 31 U(x) U(x) = U 0 E 0 L x Figure 4.2: The semi-infinite square well potential energy function. the lowest energies, (see graphs of φ 1 (x) and φ 2 (x)) the wavefunctions are mostly confined to x < 0 < L/2, but there is non-zero probability that the particle can be found in the classically forbidden region to the right of the midpoint. (According to classical mechanics the particle would have to have negative kinetic energy to be in this region!) This is an example of a phenomenon called quantum tunneling. We will investigate tunneling further in the next subsection. The illustrated wavefunctions are all special-case wave functions with definite energies. Unlike the case of a particle in box, there is no formula, analogous to Eq. (3.9), for the energies of these states. These energies can be determined numerically with a computer using methods like those you will explore in lab The semi-infinite square well and quantum tunneling We now turn to in the potential illustrated in Fig. 4.2, known as the semiinfinite square well potential. This is very similar to the more complicated box introduced in the previous subsection. Within the region 0 < x < L the potential U(x) is zero, just like it was for the particle-in-box problem. The solution of Schrödinger s equation in that region is exactly as it was for the particle in a box, up to the point that we had determined that wavefunction must have the form ψ(x) = A sin(kx), (4.35) where k 2 = 2mE/ h 2. But in this case the wavefunction doesn t necessarily go to zero at x = L, so we can t use the condition ψ(l) = 0 to determine values of k.

10 32 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION In the region x > L, the potential energy is a constant U(x) = U 0, and Schrödinger s equation, Eq. (4.1), becomes and a slight rearrangement yields h2 d 2 ψ(x) 2m dx 2 + U 0 ψ(x) = Eψ(x), (4.36) d 2 ψ(x) dx 2 = [ 2m h 2 (U 0 E) ] ψ(x). (4.37) In what follows we will focus on situations in which U 0 > E, as illustrated in Fig The constant within the square brackets on the right is a positive number. For convenience we label this constant κ 2, where and Schrödinger s equation becomes κ 2 = 2m(U 0 E) h 2 > 0, (4.38) d 2 ψ(x) dx 2 = κ 2 ψ(x). (4.39) Our job is to determine the function ψ(x) that is a solution of this differential equation. The question is: What function has a second derivative that is equal to the function itself times a positive number? There are only two real functions that have this property: Ce κx and Ce κx. (If it isn t obvious that these are solutions, use the guess-and-check to convince yourself.) So outside the box, the solution must be of the form ψ(x) = Ce κx + De κx, (4.40) where the constants C, D, and κ, have yet to be determined. The condition that the wavefunction must be normalizable imposes the constraint that D = 0; if it wasn t, ψ would grow infinitely large at large values of x, and we couldn t make the area under ψ 2 be 1 with any normalization constant. So we re down to ψ(x) = Ce κx for x > L, (4.41) where κ = 2m(U 0 E)/ h. Stop for a moment and think about what Schrödinger s equation is telling us. The region where x > L is classically forbidden, but the wavefunction isn t zero there. That means there is a non-zero probability density for the particle to be located in the classically forbidden region! The total wavefunction is thus of the form 0 for x < 0 ψ(x) = A sin (kx) for 0 x L, (4.42) Ce κx for L < x

11 4.3. SOLUTIONS OF SCHRÖDINGER S EQUATION 33 ψ 1(x) ψ 2 (x) x = L x x = L x Figure 4.3: The lowest two energy states of a particle in a semi-infinite potential well. with three constants yet to be determined: E (which gives us both k and κ), A, and C. But we have three pieces of information that we haven t yet used: Wavefunctions must be continuous, which means that the wave function just to the left of x = L should match the wave function just to the right of x = L, or A sin(kl) = Ce kl. (4.43) Wavefunctions shouldn t have kinks (unless the potential jumps to infinity), which means that derivative of the wavefunction just to the left of x = L should match the derivative just to the right of x = L, or dψ = dψ (4.44) dx dx x=l Wavefunctions must be normalized. x=l + Using these three pieces of information to determine the constants k, κ, and A (and thus the energy) isn t conceptually difficult, but the algebra gets messy, and results in an equation that must be solved numerically, so we won t do that here. But in class we will discuss some computer demos of what happens when you try to fit waves into this box. You will see that once again, only certain discrete values of k (and thus only certain discrete values of E) work, in the sense that they give valid solutions of Schrödinger s equation. Graphs of the wavefunctions for the two lowest energy states ψ 1 (x) and ψ 2 (x) are shown in Fig An examination of Fig. 4.3 leaves us the following important conclusions: Wavefunctions in the region where E > U are sinusoidal functions similar to those in the infinite square well potential. The wavelength depends on the difference E U.

12 34 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION Wavefunctions in the region where E < U are exponential functions. In the case of the semi-infinite well, the length of the well L is not exactly an integer or half-integer multiple of the wavelength as in the infinite square well (see Fig.??) Particle in a harmonic oscillator potential A fourth physical system we consider is the familiar one-dimensional harmonic oscillator. This is a single particle confined to move in one dimension, say along the x-axis, by a linear restoring force F sp = k sp x, with an angular frequency ω = k/m. The potential energy for the oscillator is illustrated on the left in Fig. 4.4, and given by U sp = 1 2 k spx 2 = 1 2 mω2 x 2. (4.45) Classically, a particle confined by such a potential energy function oscillates back and forth around the minimum of the potential energy function at x = 0, and the particle can have any total energy E tot. Conservation of mechanical energy gives the amplitude of the oscillation A = 2E tot /k, and for a given energy the particle is confined to the region A < x < A. The special-case wavefunctions for a particle in a harmonic oscillator potential are solutions to Schrödinger s equation with the harmonic oscillator potential inserted: h2 d 2 ψ(x) 2m dx mω2 x 2 ψ(x) = Eψ(x), (4.46). We will not solve this equation for the special case wavefunctions that s material for a more advanced course but we will give formulas for some of the states, and you will be able to demonstrate that they are, in fact, solutions of Schrödinger s equation. Several low-energy special-case quantum wavefunctions for the harmonic oscillator are illustrated on the right in Fig As the energy corresponding to the wavefunctions increases, additional antinodes are added to the wavefunctions, as in the case of the particle in a box. But the wavefunctions are not sinusoidal, as the wavelength changes with position. Mathematical expressions for the three lowest-energy wavefunctions are given by ψ HO 0 (x) = 1 (a 2 π) [ ψ HO 1 (x) = 2 x a 1 x 2 e 2 1/4 1 (a 2 π) ψ2 HO (x) = 1 ( 2 x2 2 a 2 1 a 2 (4.47) ] 1 x 2 e 2 a 1/4 2 ) [ 1 (a 2 π) ] 1 x 2 e 2 a 1/4 2 (4.48), (4.49)

13 4.3. SOLUTIONS OF SCHRÖDINGER S EQUATION ω 3 2 ω 1 2 ω U 0 x (units: /(mω)) φ0(x) (units: 1/ 4 /(mω)) φ 1(x) (units: 1/ 4 /(mω)) φ 2(x) (units: 1/ 4 /(mω)) φ8(x) (units: 1/ 4 /(mω)) x (units: /(mω)) x (units: /(mω)) x (units: /(mω)) x (units: /(mω)) Figure 4.4: Particle in a harmonic oscillator potential.

14 36 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION where a is a constant with dimensions of length, h a mω. (4.50) The definite energies of the special-case wavefunctions are given by the simple formula ( E n = hω n + 1 ), (4.51) 2 where n is an integer, this time starting with n = 0. giving a ladder of equally spaced energy levels. As in the previous cases, the minimum energy is not zero; in this case it is E min = E 0 = 1 2 hω. (4.52) Once again, we can t fit a wave with less than one antinode in the confining potential. This time we have both some kinetic energy and some potential energy associated with the minimum energy wavefunction. The classical turning points for the energies associated with the wavefunctions are indicated with vertical dashed lines in both the potential energy diagram and in the wavefunction graphs. Once again we see a non-zero wavefunction, and thus a non-zero probability to find the particle in a classically forbidden region. Example 4. First test of a solution for a particle in a harmonic oscillator potential You have seen that sinusoidal waves work for the particle in a box problem. Try the test function ψ(x) = A sin(kx) to see if a function of this form can be a wavefunction for a particle in a harmonic oscillator potential. Solution: To use Schrödinger s equation, we need second derivative of the wavefunction: d 2 ψ dx 2 = Ak2 sin(kx) (4.53) Inserting this into Schrödinger s equations gives k 2 h 2 2m A sin(kx) mω2 x 2 A sin(kx)? = E A sin(kx) (4.54) After canceling common factors we are left with k 2 h 2 2m mω2 x 2? = E (4.55)

15 4.3. SOLUTIONS OF SCHRÖDINGER S EQUATION 37 There are no choices we can make for k and E that will make this true for all values of x we can t make two constant terms add up to equal a term with an x 2. (Choosing a cosine for ψ would lead to the same result.) Conclusion: sinusoidal waves don t fit in parabolic boxes. Example 5. Second test of a solution for a particle in a harmonic oscillator potential Try the test function ψ(x) = xe bx2 to see if a function of this form can be a wavefunction for a particle in a harmonic oscillator potential. If it is a possible wavefunction, determine the unknown constant a in the solution and the unknown energy E. Solution: Again, we must calculate the second derivative of this function. Although ψ(x) is complicated, we will leave the calculation of the second derivative for you to do as a problem (see problem 15 at the end of this chapter). The final result for the second derivative is d 2 ψ 2 (x) dx 2 = ( 4b 2 x 3 6bx ) e bx2. (4.56) Substituting this and ψ(x) into the Schrödinger equation, Eq. (??), gives h2 ( 4b 2 x 3 6bx ) e bx2 + 1 ( 2m 2 mω2 x 2 x e bx2) ( = E x e bx2). (4.57) Canceling common factors simplifies this to 2 h 2 b 2 m and rearrangement gives x 2 ( 1 2 mω2 2 h2 b 2 x2 + 3 b h2 m mω2 x 2 = E, (4.58) m ) ) + (3 b h2 m E = 0. (4.59) The only way to make this true for all values of x is for both of the constant factors in parentheses to independently be zero: 1 2 mω2 2 h2 b 2 m Solving the first equation for b gives = 0 and 3b h2 m b = mω 2 h E = 0. (4.60) (4.61)

16 38 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION and the second for E (and using the result for b) gives E = 3 b h2 m ( mω ) h 2 = 3 2 h m 3 = (4.62) 2 hω. We see that ψ(x) = xe bx2 is a solution to the Schrödinger equation if the constants b and E are as given in the previous equation. We see that the Schrödinger equation gives us the energy E for this state, and also determines the unknown constant b in our trial wavefunction, ψ. This wavefunction solution is actually just an unnormalized version of wavefunction for the first excited state of the quantum oscillator that was given as φ HO 2 (x) in Eq. (4.48), with definite energy E 2 = 2 hω. 3 In one of the assigned problems at the end of the chapter, you will test the wavefunction solution corresponding to the ground state of the harmonic oscillator. 4.4 Heisenberg s Uncertainty Principle One of the consequences of accepting de Broglie s proposal that matter exhibits wave-like behavior is that we must give up on the notion that particles have definite positions. Another way to say this is that there is an inherent uncertainty in the position of a particle. There will also be an inherent uncertainty in the momentum of a particle. We use the special-case wavefunctions of a particle in a box (illustrated in Fig. 3.1) to explore the relationship between these two uncertainties. One way to specify the position of the particle in a box is to say somewhere between x = 0 and x = L. Another way would be to say x = 0.5L ± 0.5L, meaning that the probability distribution is centered at 0.5L with a spread of values, or uncertainty, that is also 0.5L. There is a more precise way to characterize the uncertainty, which is the standard deviation of the probability distribution: σ x = x 2 x 2 (4.63) (which should look familiar from lab). But for now we re going to use the simple (over)estimate σ x L 2. (4.64) As discussed earlier, one approximate way to think about the specialcase wavefunctions is as two counter-propagating traveling waves. In the

17 4.4. HEISENBERG S UNCERTAINTY PRINCIPLE 39 n = 3 wavefunction illustrated at the top of Fig. 3.1, the wavelength is 2L/3, corresponding to the two momenta p 1 = + h λ = 3h 2L and p 2 = h λ = 3h 2L. (4.65) One way to characterize our knowledge of the momentum is p = 0 ± 3h 2L, (4.66) or to say that our uncertainty in the momentum is σ p = 3h/(2L). If we take the product of the two uncertainties we see that the factors of L cancel: σ x σ p L 3h 2 2L = 3 h. (4.67) 4 Is there a situation in which the product of these uncertainties is smaller, so that things aren t quite so non-classical? One answer is clear in Fig. 3.1: go to the lower energy states, where σ x is still about the same, but the wavelength is longer, and therefore the uncertainty in momentum is smaller. For the lowest-energy state we find σ x σ p L 2 h 2L = 1 h. (4.68) 4 Can we do any better by making L smaller, and reducing σ x, our uncertainty in the position? Only at cost of increasing σ p, our uncertainty in the momentum by the same factor; we have reached a minimum in the product of our uncertainties in the lowest-energy wavefunction. One way to express this is σ x σ p 1 4 h = π h, APPROXIMATE (4.69) 2 where we have introduced the symbol h h 2π (because physicist s get tired of writing all the 2π s). A more sophisticated computation of σ x and σ p as standard deviations of probability distributions changes the factor of π in the numerator to 1, yielding the famous Heisenberg uncertainty principle: σ x σ p h 2, (4.70) This expresses a fundamental limit on our ability to characterize quantum particles with classical concepts. Its origins lie in the wave nature of particulate matter. It says that we cannot simultaneously know a particle s position and momentum, because a precise position and a precise momentum would mean both σ x and σ p are zero, but that is not possible. Put another way, whenever a particle is confined, i.e., if σ x is less than, then there is a minimum spread σ p 0 in the momentum. Assume

18 40 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION the product σ x σ p in the Heisenberg uncertainty relation, Eq. (??), to be as small as possible, namely, σ x σ p h/2. This gives us the inverse relation σ p h 2σ x. (4.71) If σ p 0, there is a non-zero kinetic energy. The average momentum must again be zero by symmetry, p = 0, so σ 2 p = p 2 0 p 2 = p 2. (4.72) The average kinetic energy can be related to the spread in momentum: The average kinetic energy can be related to the spread in momentum: K = 1 2 m v2 = 1 2m p2 = σ2 p 2m. (4.73) Conceptually, the since the electron is confined, it is forced to have some spread in velocity or momentum. The larger this spread, the larger the resulting average kinetic energy. Recall that when we solved for the motion of an object in classical mechanics, say a blow dart shot straight upward, we assumed that it was possible to specify the initial position of the dart and the initial velocity, but the Heisenberg uncertainty principle says that is not possible with infinite precision. Is there a contradiction? Not on the macroscopic scale, where the size of h means that precision limits imposed by Heisenberg are well beyond any possibility of detection. But down at the atomic scale the Heisenberg uncertainty relation has a very big impact. And, of course, experiments have confirmed the uncertainty relation, so it is not just a proposal but it is part of reality. Many people, when encountering the uncertainty principle for the first time, assume that this is a statement about the limits of our ability to do experiments. But that is not the case. It s a property of nature obeyed by all matter, even when we re not looking.

19 PROBLEMS 41 Problems 1. [4.6] The potential energy for the one-dimensional finite square well is shown in Fig. 4.5, with dotted lines representing the energies of the ground state and the first excited state. U(x) E excited E ground 0 L x Figure 4.5: Plot of U(x) versus x, for Problem 1. (a) Using general principles developed in Section and illustrated in Fig. 4.1 on p. 30, sketch the ground state wavefunction versus position, including regions in which E < U. (b) Compared to the ground state wavefunction of the infinite square well with the same width, is the wavelength in the classicallyallowed region longer or shorter? Is the energy larger or smaller? (c) Repeat (a) and (b) for the first excited state. (d) Note that the wavefunction isn t zero for x < 0 or for x > L, so there is a non-zero probability that the particle could be located in these regions. Why is this a violation of classical physics? (Hint: what can you say about the kinetic energy and speed of the particle when it is in one of these regions?) 2. Calculate the probability that a measurement of mass on a spring oscillating in the n = 0 state will yield a position that is classically forbidden. Hint 1: Remember the result of Problem 18. Hint 2: A change of variables may come in handy.

20 42 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION 3. Consider the illustrated wavefunction. Sketch a potential U(x) that would result in a wavefunction like this. Figure 4.6: Figure for problem 3 4. The wavefunction solution to the semi-infinite square well was given in Eq. (4.42) for the two regions 0 < x < L and x > L: ψ(x) = { A sin (kx) for 0 < x < L De κx for x > L. (4.74) In order for the wavefunction solution to be continuous and smooth across the boundary at x = L, the value of the wavefunction and its first derivative for the two solutions must match up at the boundary at x = L. (a) Using the solutions given in Eq. (4.42), write an equation that states that the two wavefunction solutions are equal at the boundary x = L. (Your equation should only contain the symbols k, κ, A, D and numerical constants.) (b) Using the solutions given in Eq. (4.42), write an equation that states that the d erivatives of the two wavefunction solutions are equal at the boundary x = L. (Your equation should only contain the symbols k, κ, A, D and numerical constants.) (c) Using the Schrödinger equation, it can be shown that the quantities k and κ are related to the energy E of the particle according to k 2 = 2mE h 2 and κ 2 = 2m (U 0 E) h 2 (4.75) where m is the mass of the particle and U o is the height of the potential well. Given these relations and the results of parts (a) and (b), do you think that the particle can have any value of energy E in the well?

21 PROBLEMS Semi-infinite square-well potential. Download the Excel worksheet semi-finite.xls from either the Handouts page or from the Calendar page. This sheet shows the calculations for determining the wavefunctions for a potential well that is infinite at x = 0 but of finite magnitude on the right side of the well (which is at x = 5 in this problem). You ll see two graphs: the top one shows the semi-infinite potential well (in purple) along with a non-normalized plot of the calculated wavefunction so you can see it along with the potential. The bottom graph shows the normalized wavefunction, corresponding to the second-to-last column in the worksheet. When you bring up the worksheet, the energy will be set for the value for the ground state. Some questions: (a) Sketch or print out (just the first page!) the wavefunctions that are displayed for the ground state along with at least two of the excited states. To display the 1 st and 2 nd excited states, type in and respectively in the framed box for energy. (b) What happens if you type in an energy that isn t one of the welldefined energies for the problem? Try it out, and comment on what happens. Had we not told you what the allowed energies were, how might you figure them out? (You ll be doing this in lab later this semester.) 6. [4.10] Given the solution to Schrödinger s equation for an electron in the semi-infinite square well potential (see Section 4.3.3). Assume that U 0 = 12 ev and the electron is in a state with an energy E = 10 ev. (a) Calculate the value of the decay constant κ for the wavefunction in the classically-forbidden region. (b) Calculate the distance into the classically-forbidden region beyond x = L where the probability density is a factor of 2 smaller than that at x = L (i.e., calculate the value of d such that ψ(l + d) 2 / ψ(l) 2 = e 2κd = 0.5. (c) Now, let s assume that the particle is a ball with mass 0.5 kg, and assume that U 0 and E have everyday values of, say, 100 J and 50 J, respectively. Repeat the calculations from parts (a) and (b). (d) Why do you think we never experience quantum tunneling for everyday objects? 7. Consider a proton (m p = kg) confined in an infinite potential well of size L = m. Calculate the energy of a photon emitted when the proton makes a transition from the n = 4 state to the n = 3 state.

22 44 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION 8. Electrons in an ensemble of 10 nm wide one-dimensional quantum dots are all initially in the n = 4 state. Find the wavelengths of all possible photons emitted as electrons make transitions to the ground state. 9. Calculate the width of a one-dimensional quantum dot such that a transition of a trapped electron from the first excited state (n = 2) to the ground state (n = 1) will produce red light with a wavelength of 630 nm. 10. (a) Ozone (O 3 ) in the atmosphere absorbs ultraviolet radiation that can damage the skin and cause cancer. It does this via a biochemistry process O 3 + γ O 2 + O, where γ is a photon. If the dissociation energy of ozone is 3.94 ev, calculate the maximum wavelength of UV radiation that is absorbed by this photo-chemical process. (b) Diatomic oxygen (O 2 ) also absorbs UV radiation with smaller wavelengths via the process O 2 +γ O+O. Given a dissociation energy of 5.17 ev for an O 2 molecule, calculate the maximum wavelength of UV radiation that is absorbed by this process. 11. Titanium dioxide is one of many different possible ingredients used in sunblock lotions. It has a band gap energy of 3.2 ev. Calculate the maximum wavelength of radiation that you would expect TiO 2 to absorb via an absorption process that promotes electrons from the valence to the conduction band. Is your answer consistent with the use of TiO 2 as an ingredient in sunblock lotions? 12. Carbon dioxide has first and second excited vibrational states which are ev and ev above the ground state. Calculate the wavelengths of electromagnetic radiation that you would expect to be most readily absorbed by transitions between these vibrational energy levels in CO 2. The wavelengths that you will find are in the infrared range of 700 nm to 1 mm. This is one of the reasons why carbon dioxide contributes to global warming as IR radiation is one of the ways in which the Earth radiates heat, assuming it isn t absorbed in the atmosphere. (Note, though, that this is a simplification, as there are rotational energy levels in CO 2 as well that aren t included in this problem.) 13. Class 3 (neutral phenol) Green Fluorescence Protein (GFP) used in cell and molecular biology studies has an excitation wavelength of 399 nm (near-uv) and an emission wavelength of 511 nm (green, duh). (a) Determine a 3-level energy diagram (with ground state energy 0 ev) that is consistent with these data. (There are two answers that are consistent with the given data.) (b) Based on your diagram, you might expect to see two different colors of emitted light. Why do you see only green light emitted from a sample tagged (or genetically mutated) with Class 3 GFP?

23 PROBLEMS In Section we demonstrated that there were two possible solutions to Schrödinger s equation in tunneling regions in which E < U 0, where U 0 is a constant: A 1 e κx and A 2 e κx. We then asserted that the sum ψ(x) = A 1 e κx + A 2 e κx is also a solution of Schrödinger s equation. Show that this is true. 15. In Example 5 we needed to use the second derivative of the wavefunction for the first excited state of the harmonic oscillator, φ HO 1 (x). Calculate this derivative. Use any technological assistance that makes this easy. 16. [3.16] A special case of the quantum harmonic oscillator is described by the Schrödinger equation of the form d2 ψ dx 2 + 4x2 ψ = Eψ. Substitute the trial solution ψ = e x2 energy E. and determine the value of the 17. [3.17] The general case Schrödinger equation for the one-dimensional harmonic oscillator is h2 d 2 ψ 2m dx mω2 x 2 ψ = Eψ. Substitute the trial solution ψ(x) = e bx2 into this equation and determine the constants b and E. This is the ground state of the oscillator. 18. Show that the classical turning points for a harmonic oscillator with the energy E n = (n ) hω are given by A n = 2n + 1 a, where a is the constant used in the harmonic oscillator wavefunctions, a = h/(mω). Compare this result with the positions of the dashed vertical lines in Fig Consider an iodine atom with atomic mass 127 that is bound by a springlike force with k sp = 350 nn/nm = 350 N/m. (This not crazy at all atoms are bound to each other in molecules that vibrate.) (a) Calculate the classical oscillation frequency of the atom. (b) Use the result of Problem 18 to determine the amplitude of the classical oscillation with the lowest possible quantum energy.

24 46 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION 20. Calculate the probability that a measurement of mass on a spring oscillating in the n = 0 state will yield a position that is classically forbidden. Hint 1: Remember the result of Problem 18. Hint 2: A change of variables may come in handy. 21. The figure shows the graph of a wavefunction ψ(x) and the associated probability density ( ψ 2 ) for a particle. Estimate the probability that a measurement of the position of the particle will yield a result that is classically forbidden. 2 1 ψ x ψ x Figure 4.7: Figure for problem 21.

25 PROBLEMS Consider a particle mass on spring in the quantum state with the third lowest energy, i.e., with n = 2. Throughout this problem you may assume that you are using units such that a = h/mω = 1. (a) Make a graph of the probability density for position of this particle. (b) Demonstrate that the wavefunction for this particle is normalized. (c) Calculate the probability that a measurement of the position of the particle will yield a result that is classically allowed. 23. [3.5] An electron in a hydrogen atom has a spread in position typically around σ x = m. (a) Use the Heisenberg uncertainty relation to find a lower bound on the spread in momentum, σ px, for the electron. (b) Now find a lower bound on the spread in velocity, σ vx. 24. [3.6] The Heisenberg uncertainty relation does not pose much of a limitation on our macroscopic scale. Consider a blow dart of mass 2.5 g. Let s imagine a measurement of the dart s position with a precision limited to 1 µm. Determine the lower bound on the spread in velocity imposed by the Heisenberg uncertainty relation. 25. [3.7] Fabrication of nano-scale devices requires the ability to position atoms and molecules with very small spatial spread σ x. (a) Use Heisenberg s uncertainty principle to make a rough estimate of the precision by which a carbon atom (mass kg and radius 70 pm) can be confined (i.e., determine the smallest σ x ) to keep its minimum kinetic energy below typical molecular binding energies of around 1 ev. Would you expect the uncertainty principle to be a problem if you wanted to arrange carbon atoms in a nanotech device with a precision of around one-hundredth of the radius of a carbon atom? (b) Repeat the calculations in part (a) for a proton (mass kg), also with K below 1 ev. (c) Compare your result from (b) to the size of the nucleus of a carbon atom ( m). What does this imply about a proton confined inside a carbon nucleus? I.e., what must be true for a proton to remain bound inside a carbon nucleus? 26. [3.8] Back in 1975, Gordon Moore proposed that the number of transistors per area on integrated circuits roughly doubles every two years. This principle ( Moore s Law ) has worked surprisingly well for 40 years now,

26 48 CHAPTER 4. SCHRÖDINGER S WAVE EQUATION with transistors introduced in 2012 as small as 22 nm and techniques are continually being developed to make them even smaller. But Moore s Law will eventually fail due to limitations imposed by the uncertainty principle. (a) Use the uncertainty principle to calculate the smallest spread σ x for an electron such that its minimum kinetic energy (due to uncertainty) is below the work function (binding energy) of silicon, which is 4.05 ev. (b) Let s assume that the smallest possible transistor has an area 100 times the square of the σ x that you calculated in part (a). Given the area (approximated as the square of 22 nm) for the best transistors from 2012, if Moore s Law continues to hold into the future, roughly what year will transistors reach this quantum limit?